On a pulley of mass M hangs a rope with two m | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Since the pulley is massive,the tensions on both sides of the pulley are not equal. From the free body diagram,we have:$m_{1} g - T_{1} = m_{1} a$

Vedclass · Accepted Answer

Vedclass · Answer

(C) Since the pulley is massive,the tensions on both sides of the pulley are not equal. From the free body diagram,we have:$m_{1} g - T_{1} = m_{1} a$  ---$(i)$$T_{2} - m_{2} g = m_{2} a$  ---(ii)$(T_{1} - T_{2}) R = I \alpha = \left(\frac{M R^{2}}{2}\right) \left(\frac{a}{R}\right) = \frac{M R a}{2}$  ---(iii)From equation (iii),we get:$T_{1} - T_{2} = \frac{M a}{2}$  ---(iv)Substituting $T_{1} = m_{1}(g - a)$ and $T_{2} = m_{2}(g + a)$ into equation (iii):$m_{1}(g - a) - m_{2}(g + a) = \frac{M a}{2}$$(m_{1} - m_{2}) g = a \left(\frac{M}{2} + m_{1} + m_{2}\right)$$a = \frac{(m_{1} - m_{2}) g}{\frac{M}{2} + (m_{1} + m_{2})}$Substituting $a$ into equation (iv):$T_{1} - T_{2} = \frac{M}{2} \cdot \frac{(m_{1} - m_{2}) g}{\frac{M}{2} + (m_{1} + m_{2})} = \frac{(m_{1} - m_{2}) g}{1 + \frac{2(m_{1} + m_{2})}{M}}$As $M \rightarrow 0$,$T_{1} - T_{2} \rightarrow 0$. As $M \rightarrow \infty$,$T_{1} - T_{2} \rightarrow (m_{1} - m_{2}) g$. The expression shows that the difference increases with $M$ and approaches a constant value,which corresponds to the curve in graph $(c)$.

Column-$I$	Column-$II$
$(1)$ Newton's third law of motion	$(a)$ $\vec{F}_{12} = -\vec{F}_{21}$
$(2)$ Law of conservation of momentum	$(b)$ $\Delta \vec{p} = 0$
	$(c)$ $\vec{F}_{12} = \vec{F}_{21}$

Similar Questions

Imagine a situation in which the horizontal surface of block $M_0$ is smooth and its vertical surface is rough with a coefficient of friction $\mu$. In the above problem,choose the correct value$(s)$ of $F$ for which the blocks $M$ and $m$ remain stationary with respect to $M_0$.

Match the items in Column-$I$ with those in Column-$II$.

Column-$I$ Column-$II$

$(1)$ Newton's third law of motion $(a)$ $\vec{F}_{12} = -\vec{F}_{21}$

$(2)$ Law of conservation of momentum $(b)$ $\Delta \vec{p} = 0$

$(c)$ $\vec{F}_{12} = \vec{F}_{21}$

Which of the following statements is not true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module